一道二次根式的运算题求证:3/(√6+√3)+1/(√2+√3)=√6-√2

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 22:57:05
一道二次根式的运算题求证:3/(√6+√3)+1/(√2+√3)=√6-√2一道二次根式的运算题求证:3/(√6+√3)+1/(√2+√3)=√6-√2一道二次根式的运算题求证:3/(√6+√3)+1

一道二次根式的运算题求证:3/(√6+√3)+1/(√2+√3)=√6-√2
一道二次根式的运算题
求证:
3/(√6+√3)+1/(√2+√3)=√6-√2

一道二次根式的运算题求证:3/(√6+√3)+1/(√2+√3)=√6-√2
左边的两个根式先把分母有理化
3/(√6+√3)上下同乘√6-√3
1/(√2+√3)上下同乘√3-√2
左边得3*(√6-√3)/(6-3)+(√3-√2)/(3-2)
化简就等于右边

3=(√6+√3)*(√6-√3),1/(√2+√3)=√3-√2.这就行了